The least value of n for which the number of integral terms in the Binomial expansion of (\sqrt[3]7+\sqrt[12]11)^n is 183, is :

Binomial Theorem PYQ — JEE Main 2025 · 29 Jan · Shift 1 · Q34 with Solution

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Correct answer: A

General term of the expansion

For $\left(7^{1/3}+11^{1/12}\right)^n,$ the general term is

=\binom{n}{r}7^{\frac{n-r}{3}}11^{\frac{r}{12}}.$$ For this term to be **integral**, both exponents must be integers: - $$\frac{n-r}{3}\in \mathbb Z$$ - $$\frac{r}{12}\in \mathbb Z$$ So we need $$r\equiv 0\pmod{12}$$ and $$n-r\equiv 0\pmod{3}.$$ But if $r$ is a multiple of $12$, then it is automatically a multiple of $3$. Hence $n-r\equiv 0\pmod 3$ becomes $$n\equiv 0\pmod 3.$$ Thus: - if $n$ is **not** divisible by $3$, there are no integral terms; - if $n$ **is** divisible by $3$, then every $r=0,12,24,\dots\le n$ gives an integral term. --- 2. **Count the number of integral terms** When $3\mid n$, the valid values of $r$ are: $$r=0,12,24,\dots,12\left\lfloor \frac{n}{12}\right\rfloor.$$ Hence the number of integral terms is $$\left\lfloor \frac{n}{12}\right\rfloor+1.$$ We are given this number equals $183$: $$\left\lfloor \frac{n}{12}\right\rfloor+1=183.$$ So, $$\left\lfloor \frac{n}{12}\right\rfloor=182.$$ Therefore, $$2184\le n<2196.$$ Also, we need $n\equiv 0\pmod 3$. The least multiple of $3$ in the interval $[2184,2196)$ is $$n=2184.$$ --- 3. **Check options** - **A: 2184** $$\left\lfloor \frac{2184}{12}\right\rfloor+1=182+1=183,$$ and $2184$ is divisible by $3$. Correct. - **B: 2172** $$\left\lfloor \frac{2172}{12}\right\rfloor+1=181+1=182.$$ Not correct. - **C: 2196** $$\left\lfloor \frac{2196}{12}\right\rfloor+1=183+1=184.$$ Not correct. - **D: 2148** $$\left\lfloor \frac{2148}{12}\right\rfloor+1=179+1=180.$$ Not correct. --- 4. **Final answer** The least value of $n$ is $$\boxed{2184}.$$

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View written solutionFree

Correct answer: A

General term of the expansion

For $\left(7^{1/3}+11^{1/12}\right)^n,$ the general term is

=\binom{n}{r}7^{\frac{n-r}{3}}11^{\frac{r}{12}}.$$ For this term to be **integral**, both exponents must be integers: - $$\frac{n-r}{3}\in \mathbb Z$$ - $$\frac{r}{12}\in \mathbb Z$$ So we need $$r\equiv 0\pmod{12}$$ and $$n-r\equiv 0\pmod{3}.$$ But if $r$ is a multiple of $12$, then it is automatically a multiple of $3$. Hence $n-r\equiv 0\pmod 3$ becomes $$n\equiv 0\pmod 3.$$ Thus: - if $n$ is **not** divisible by $3$, there are no integral terms; - if $n$ **is** divisible by $3$, then every $r=0,12,24,\dots\le n$ gives an integral term. --- 2. **Count the number of integral terms** When $3\mid n$, the valid values of $r$ are: $$r=0,12,24,\dots,12\left\lfloor \frac{n}{12}\right\rfloor.$$ Hence the number of integral terms is $$\left\lfloor \frac{n}{12}\right\rfloor+1.$$ We are given this number equals $183$: $$\left\lfloor \frac{n}{12}\right\rfloor+1=183.$$ So, $$\left\lfloor \frac{n}{12}\right\rfloor=182.$$ Therefore, $$2184\le n<2196.$$ Also, we need $n\equiv 0\pmod 3$. The least multiple of $3$ in the interval $[2184,2196)$ is $$n=2184.$$ --- 3. **Check options** - **A: 2184** $$\left\lfloor \frac{2184}{12}\right\rfloor+1=182+1=183,$$ and $2184$ is divisible by $3$. Correct. - **B: 2172** $$\left\lfloor \frac{2172}{12}\right\rfloor+1=181+1=182.$$ Not correct. - **C: 2196** $$\left\lfloor \frac{2196}{12}\right\rfloor+1=183+1=184.$$ Not correct. - **D: 2148** $$\left\lfloor \frac{2148}{12}\right\rfloor+1=179+1=180.$$ Not correct. --- 4. **Final answer** The least value of $n$ is $$\boxed{2184}.$$

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